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2. publishing

Chapter 7
Linear Programming Models:
Graphical and Computer
Methods
To accompany
Quantitative Analysis for Management, Eleventh Edition,
by Render, Stair, and Hanna
Power Point slides created by Brian Peterson
Learning Objectives
After completing this chapter, students will be able to:
1. Understand the basic assumptions and
properties of linear programming (LP).
2. Graphically solve any LP problem that has
only two variables by both the corner point
and isoprofit line methods.
3. Understand special issues in LP such as
infeasibility, unboundedness, redundancy,
and alternative optimal solutions.
4. Understand the role of sensitivity analysis.
5. Use Excel spreadsheets to solve LP
problems.
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7-2
Chapter Outline
7.1 Introduction
7.2 Requirements of a Linear Programming
Problem
7.3 Formulating LP Problems
7.4 Graphical Solution to an LP Problem
7.5 Solving Flair Furniture’s LP Problem using
QM for Windows and Excel
7.6 Solving Minimization Problems
7.7 Four Special Cases in LP
7.8 Sensitivity Analysis
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7-3
Introduction
 Many management decisions involve
trying to make the most effective use of
limited resources.
 Linear programming (LP) is a widely used
mathematical modeling technique
designed to help managers in planning and
decision making relative to resource
allocation.
 This belongs to the broader field of
mathematical programming.
 In this sense, programming refers to modeling
and solving a problem mathematically.
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7-4
Requirements of a Linear
Programming Problem
 All LP problems have 4 properties in common:
1. All problems seek to maximize or minimize some
quantity (the objective function).
2. Restrictions or constraints that limit the degree to
which we can pursue our objective are present.
3. There must be alternative courses of action from which
to choose.
4. The objective and constraints in problems must be
expressed in terms of linear equations or inequalities.
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7-5
Basic Assumptions of LP
 We assume conditions of certainty exist and

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
numbers in the objective and constraints are
known with certainty and do not change during
the period being studied.
We assume proportionality exists in the objective
and constraints.
We assume additivity in that the total of all
activities equals the sum of the individual
activities.
We assume divisibility in that solutions need not
be whole numbers.
All answers or variables are nonnegative.
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7-6
LP Properties and Assumptions
PROPERTIES OF LINEAR PROGRAMS
1. One objective function
2. One or more constraints
3. Alternative courses of action
4. Objective function and constraints are linear
– proportionality and divisibility
5. Certainty
6. Divisibility
7. Nonnegative variables
Table 7.1
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7-7
Formulating LP Problems
 Formulating a linear program involves developing
a mathematical model to represent the managerial
problem.
 The steps in formulating a linear program are:
1. Completely understand the managerial
problem being faced.
2. Identify the objective and the constraints.
3. Define the decision variables.
4. Use the decision variables to write
mathematical expressions for the objective
function and the constraints.
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7-8
Formulating LP Problems
 One of the most common LP applications is the
product mix problem.
 Two or more products are produced using
limited resources such as personnel, machines,
and raw materials.
 The profit that the firm seeks to maximize is
based on the profit contribution per unit of each
product.
 The company would like to determine how
many units of each product it should produce
so as to maximize overall profit given its limited
resources.
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7-9
Flair Furniture Company
 The Flair Furniture Company produces
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inexpensive tables and chairs.
Processes are similar in that both require a certain
amount of hours of carpentry work and in the
painting and varnishing department.
Each table takes 4 hours of carpentry and 2 hours
of painting and varnishing.
Each chair requires 3 of carpentry and 1 hour of
painting and varnishing.
There are 240 hours of carpentry time available
and 100 hours of painting and varnishing.
Each table yields a profit of $70 and each chair a
profit of $50.
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7-10
Flair Furniture Company Data
The company wants to determine the best
combination of tables and chairs to produce to reach
the maximum profit.
HOURS REQUIRED TO
PRODUCE 1 UNIT
DEPARTMENT
(T)
TABLES
(C)
CHAIRS
AVAILABLE HOURS
THIS WEEK
Carpentry
4
3
240
Painting and varnishing
2
1
100
$70
$50
Profit per unit
Table 7.2
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7-11
Flair Furniture Company
 The objective is to:
Maximize profit
 The constraints are:
1. The hours of carpentry time used cannot
exceed 240 hours per week.
2. The hours of painting and varnishing time
used cannot exceed 100 hours per week.
 The decision variables representing the actual
decisions we will make are:
T = number of tables to be produced per week.
C = number of chairs to be produced per week.
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7-12
Flair Furniture Company
 We create the LP objective function in terms of T
and C:
Maximize profit = $70T + $50C
 Develop mathematical relationships for the two
constraints:
 For carpentry, total time used is:
(4 hours per table)(Number of tables produced)
+ (3 hours per chair)(Number of chairs produced).
 We know that:
Carpentry time used ≤ Carpentry time available.
4T + 3C ≤ 240 (hours of carpentry time)
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7-13
Flair Furniture Company
 Similarly,
Painting and varnishing time used
≤ Painting and varnishing time available.
2 T + 1C ≤ 100 (hours of painting and varnishing time)
This means that each table produced
requires two hours of painting and
varnishing time.
 Both of these constraints restrict production
capacity and affect total profit.
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7-14
Flair Furniture Company
The values for T and C must be nonnegative.
T ≥ 0 (number of tables produced is greater
than or equal to 0)
C ≥ 0 (number of chairs produced is greater
than or equal to 0)
The complete problem stated mathematically:
Maximize profit = $70T + $50C
subject to
4T + 3C ≤ 240 (carpentry constraint)
2T + 1C ≤ 100 (painting and varnishing constraint)
T, C ≥ 0
(nonnegativity constraint)
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Graphical Solution to an LP Problem
 The easiest way to solve a small LP
problems is graphically.
 The graphical method only works when
there are just two decision variables.
 When there are more than two variables, a
more complex approach is needed as it is
not possible to plot the solution on a twodimensional graph.
 The graphical method provides valuable
insight into how other approaches work.
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7-16
Graphical Representation of a
Constraint
Quadrant Containing All Positive Values
C
100 –
Number of Chairs
–
This Axis Represents the Constraint T ≥ 0
80 –
–
60 –
–
40 –
This Axis Represents the
Constraint C ≥ 0
–
20 –
–
Figure 7.1
|–
0
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20
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60
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80
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100
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T
Number of Tables
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7-17
Graphical Representation of a
Constraint
 The first step in solving the problem is to
identify a set or region of feasible
solutions.
 To do this we plot each constraint
equation on a graph.
 We start by graphing the equality portion
of the constraint equations:
4T + 3C = 240
 We solve for the axis intercepts and draw
the line.
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7-18
Graphical Representation of a
Constraint
 When Flair produces no tables, the
carpentry constraint is:
4(0) + 3C = 240
3C = 240
C = 80
 Similarly for no chairs:
4T + 3(0) = 240
4T = 240
T = 60
 This line is shown on the following graph:
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7-19
Graphical Representation of a
Constraint
Graph of carpentry constraint equation
C
100 –
Number of Chairs
–
80 –
(T = 0, C = 80)
–
60 –
–
40 –
–
(T = 60, C = 0)
20 –
–
Figure 7.2
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0
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20
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Number of Tables
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T
7-20
Graphical Representation of a
Constraint
Region that Satisfies the Carpentry Constraint
C
 Any point on or below
the constraint plot will
not violate the
restriction.
 Any point above the
plot will violate the
restriction.
100 –
Number of Chairs
–
80 –
–
60 –
–
(30, 40)
40 –
(70, 40)
–
20 –
(30, 20)
–
Figure 7.3
|–
0
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20
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40
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60
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80
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100
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T
Number of Tables
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Graphical Representation of a
Constraint
 The point (30, 40) lies on the plot and
exactly satisfies the constraint
4(30) + 3(40) = 240.
 The point (30, 20) lies below the plot and
satisfies the constraint
4(30) + 3(20) = 180.
 The point (70, 40) lies above the plot and
does not satisfy the constraint
4(70) + 3(40) = 400.
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Graphical Representation of a
Constraint
Region that Satisfies the Painting and
Varnishing Constraint
C
100 –
(T = 0, C = 100)
Number of Chairs
–
80 –
–
60 –
–
40 –
–
(T = 50, C = 0)
20 –
–
Figure 7.4
|–
0
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20
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Number of Tables
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T
7-23
Graphical Representation of a
Constraint
 To produce tables and chairs, both

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departments must be used.
We need to find a solution that satisfies both
constraints simultaneously.
A new graph shows both constraint plots.
The feasible region (or area of feasible
solutions) is where all constraints are satisfied.
Any point inside this region is a feasible
solution.
Any point outside the region is an infeasible
solution.
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Graphical Representation of a
Constraint
Feasible Solution Region for the Flair
Furniture Company Problem
C
100 –
Number of Chairs
–
80 –
Painting/Varnishing Constraint
–
60 –
–
40 –
–
Carpentry Constraint
20 – Feasible
– Region
Figure 7.5
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0
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20
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Number of Tables
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T
7-25
Graphical Representation of a
Constraint
 For the point (30, 20)
Carpentry
constraint
4T + 3C ≤ 240 hours available
(4)(30) + (3)(20) = 180 hours used

Painting
constraint
2T + 1C ≤ 100 hours available
(2)(30) + (1)(20) = 80 hours used

 For the point (70, 40)
Carpentry
constraint
4T + 3C ≤ 240 hours available
(4)(70) + (3)(40) = 400 hours used

Painting
constraint
2T + 1C ≤ 100 hours available
(2)(70) + (1)(40) = 180 hours used

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Graphical Representation of a
Constraint
 For the point (50, 5)
Carpentry
constraint
4T + 3C ≤ 240 hours available
(4)(50) + (3)(5) = 215 hours used

Painting
constraint
2T + 1C ≤ 100 hours available
(2)(50) + (1)(5) = 105 hours used

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7-27
Isoprofit Line Solution Method
 Once the feasible region has been graphed, we

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

need to find the optimal solution from the many
possible solutions.
The speediest way to do this is to use the isoprofit
line method.
Starting with a small but possible profit value, we
graph the objective function.
We move the objective function line in the
direction of increasing profit while maintaining the
slope.
The last point it touches in the feasible region is
the optimal solution.
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7-28
Isoprofit Line Solution Method
 For Flair Furniture, choose a profit of $2,100.
 The objective function is then
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
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$2,100 = 70T + 50C
Solving for the axis intercepts, we can draw the
graph.
This is obviously not the best possible solution.
Further graphs can be created using larger profits.
The further we move from the origin, the larger the
profit will be.
The highest profit ($4,100) will be generated when
the isoprofit line passes through the point (30, 40).
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Isoprofit Line Solution Method
Profit line of $2,100 Plotted for the Flair
Furniture Company
C
100 –
Number of Chairs
–
80 –
–
60 –
–
$2,100 = $70T + $50C
(0, 42)
40 –
–
(30, 0)
20 –
–
Figure 7.6
|–
0
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20
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T
7-30
Isoprofit Line Solution Method
Four Isoprofit Lines Plotted for the Flair
Furniture Company
C
100 –
Number of Chairs
–
$3,500 = $70T + $50C
80 –
–
$2,800 = $70T + $50C
60 –
–
$2,100 = $70T + $50C
40 –
–
$4,200 = $70T + $50C
20 –
–
Figure 7.7
|–
0
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20
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Number of Tables
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7-31
Isoprofit Line Solution Method
Optimal Solution to the Flair Furniture problem
C
100 –
Number of Chairs
–
80 –
Maximum Profit Line
–
60 –
Optimal Solution Point
(T = 30, C = 40)
–
40 –
–
$4,100 = $70T + $50C
20 –
–
Figure 7.8
|–
0
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20
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Number of Tables
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T
7-32
Corner Point Solution Method
 A second approach to solving LP problems
employs the corner point method.
 It involves looking at the profit at every
corner point of the feasible region.
 The mathematical theory behind LP is that
the optimal solution must lie at one of the
corner points, or extreme point, in the
feasible region.
 For Flair Furniture, the feasible region is a
four-sided polygon with four corner points
labeled 1, 2, 3, and 4 on the graph.
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Corner Point Solution Method
Four Corner Points of the Feasible Region
C
100 –
Number of Chairs
2 –
80 –
–
60 –
–
3
40 –
–
20 –
–
Figure 7.9
1 |–
0
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20
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40
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4
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60
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80
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100
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T
Number of Tables
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7-34
Corner Point Solution Method
 To find the coordinates for Point 3 accurately we have to
solve for the intersection of the two constraint lines.
 Using the simultaneous equations method, we multiply the
painting equation by –2 and add it to the carpentry equation
4T + 3C = 240
– 4T – 2C = –200
C = 40
(carpentry line)
(painting line)
 Substituting 40 for C in either of the original equations
allows us to determine the value of T.
4T + (3)(40) = 240
4T + 120 = 240
T = 30
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(carpentry line)
7-35
Corner Point Solution Method
Point 1 : (T = 0, C = 0)
Profit = $70(0) + $50(0) = $0
Point 2 : (T = 0, C = 80)
Profit = $70(0) + $50(80) = $4,000
Point 4 : (T = 50, C = 0)
Profit = $70(50) + $50(0) = $3,500
Point 3 : (T = 30, C = 40)
Profit = $70(30) + $50(40) = $4,100
Because Point 3 returns the highest profit, this is
the optimal solution.
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Slack and Surplus
 Slack is the amount of a resource
that is not used. For a less-than-orequal constraint:
 Slack = Amount of resource available –
amount of resource used.
 Surplus is used with a greater-than-
or-equal constraint to indicate the
amount by which the right hand side
of the constraint is exceeded.
 Surplus = Actual amount – minimum
amount.
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Summary of Graphical Solution
Methods
ISOPROFIT METHOD
1. Graph all constraints and find the feasible region.
2. Select a specific profit (or cost) line and graph it to find the slope.
3. Move the objective function line in the direction of increasing profit (or
decreasing cost) while maintaining the slope. The last point it touches in the
feasible region is the optimal solution.
4. Find the values of the decision variables at this last point and compute the
profit (or cost).
CORNER POINT METHOD
1. Graph all constraints and find the feasible region.
2. Find the corner points of the feasible reason.
3. Compute the profit (or cost) at each of the feasible corner points.
4. Select the corner point with the best value of the objective function found in
Step 3. This is the optimal solution.
Table 7.4
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Solving Flair Furniture’s LP Problem
Using QM for Windows and Excel
 Most organizations have access to
software to solve big LP problems.
 While there are differences between
software implementations, the approach
each takes towards handling LP is
basically the same.
 Once you are experienced in dealing with
computerized LP algorithms, you can
easily adjust to minor changes.
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Using QM for Windows


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First select the Linear Programming module.
Specify the number of constraints (non-negativity
is assumed).
Specify the number of decision variables.
Specify whether the objective is to be maximized
or minimized.
For the Flair Furniture problem there are two
constraints, two decision variables, and the
objective is to maximize profit.
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Using QM for Windows
QM for Windows Linear Programming Computer
screen for Input of Data
Program 7.1A
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Using QM for Windows
QM for Windows Data Input for Flair
Furniture Problem
Program 7.1B
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Using QM for Windows
QM for Windows Output for Flair Furniture Problem
Program 7.1C
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Using QM for Windows
QM for Windows Graphical Output for Flair Furniture
Problem
Program 7.1D
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Using Excel’s Solver Command to
Solve LP Problems
 The Solver tool in Excel can be used to find
solutions to:
 LP problems.
 Integer programming problems.
 Noninteger programming problems.
 Solver is limited to 200 variables and 100
constraints.
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Using Solver to Solve the Flair
Furniture Problem
 Recall the model for Flair Furniture is:
Maximize profit =
Subject to
$70T + $50C
4T + 3C ≤ 240
2T + 1C ≤ 100
 To use Solver, it is necessary to enter formulas
based on the initial model.
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Using Solver to Solve the Flair
Furniture Problem
1. Enter the variable names, the coefficients for the
objective function and constraints, and the righthand-side values for each of the constraints.
2. Designate specific cells for the values of the
decision variables.
3. Write a formula to calculate the value of the
objective function.
4. Write a formula to compute the left-hand sides of
each of the constraints.
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Using Solver to Solve the Flair
Furniture Problem
Excel Data Input for the Flair Furniture Example
Program 7.2A
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Using Solver to Solve the Flair
Furniture Problem
Formulas for the Flair Furniture Example
Program 7.2B
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Using Solver to Solve the Flair
Furniture Problem
Excel Spreadsheet for the Flair Furniture Example
Program 7.2C
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Using Solver to Solve the Flair
Furniture Problem
 Once the model has been entered, the following
steps can be used to solve the problem.
In Excel 2010, select Data – Solver.
If Solver does not appear in the indicated place,
see Appendix F for instructions on how to
activate this add-in.
1. In the Set Objective box, enter the cell address
for the total profit.
2. In the By Changing Cells box, enter the cell
addresses for the variable values.
3. Click Max for a maximization problem and Min for
a minimization problem.
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Using Solver to Solve the Flair
Furniture Problem
4. Check the box for Make Unconstrained
Variables Non-negative.
5. Click the Select Solving Method button
and select Simplex LP from the menu that
appears.
6. Click Add to add the constraints.
7. In the dialog box that appears, enter the
cell references for the left-hand-side
values, the type of equation, and the
right-hand-side values.
8. Click Solve.
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Using Solver to Solve the Flair
Furniture Problem
Starting Solver
Figure 7.2D
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Using Solver to Solve the Flair
Furniture Problem
Solver
Parameters
Dialog Box
Figure 7.2E
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Using Solver to Solve the Flair
Furniture Problem
Solver Add Constraint Dialog Box
Figure 7.2F
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Using Solver to Solve the Flair
Furniture Problem
Solver Results Dialog Box
Figure 7.2G
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Using Solver to Solve the Flair
Furniture Problem
Solution Found by Solver
Figure 7.2H
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Solving Minimization Problems
 Many LP problems involve minimizing an objective
such as cost instead of maximizing a profit
function.
 Minimization problems can be solved graphically
by first setting up the feasible solution region and
then using either the corner point method or an
isocost line approach (which is analogous to the
isoprofit approach in maximization problems) to
find the values of the decision variables (e.g., X1
and X2) that yield the minimum cost.
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7-58
Holiday Meal Turkey Ranch
The Holiday Meal Turkey Ranch is considering
buying two different brands of turkey feed and
blending them to provide a good, low-cost diet for its
turkeys
Let
X1 = number of pounds of brand 1 feed purchased
X2 = number of pounds of brand 2 feed purchased
Minimize cost (in cents) = 2X1 + 3X2
subject to:
5X1 + 10X2 ≥ 90 ounces (ingredient constraint A)
4X1 + 3X2 ≥ 48 ounces (ingredient constraint B)
0.5X1
≥ 1.5 ounces (ingredient constraint C)
X1
≥ 0
(nonnegativity constraint)
X2 ≥ 0
(nonnegativity constraint)
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Holiday Meal Turkey Ranch
Holiday Meal Turkey Ranch data
COMPOSITION OF EACH POUND
OF FEED (OZ.)
INGREDIENT
BRAND 1 FEED
BRAND 2 FEED
MINIMUM MONTHLY
REQUIREMENT PER
TURKEY (OZ.)
A
5
10
90
B
4
3
48
C
0.5
0
Cost per pound
2 cents
1.5
3 cents
Table 7.5
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Holiday Meal Turkey Ranch
 Use the corner point method.
 First construct the feasible solution region.
 The optimal solution will lie at one of the corners
as it would in a maximization problem.
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Feasible Region for the Holiday
Meal Turkey Ranch Problem
X2
–
Pounds of Brand 2
20 –
Ingredient C Constraint
15 –
10 –
Feasible Region
a
Ingredient B Constraint
5–
Ingredient A Constraint
b
Figure 7.10
0 |–
|
5
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c
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10
15
20
Pounds of Brand 1
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Holiday Meal Turkey Ranch
 Solve for the values of the three corner points.
 Point a is the intersection of ingredient constraints
C and B.
4X1 + 3X2 = 48
X1 = 3
 Substituting 3 in the first equation, we find X2 = 12.
 Solving for point b with basic algebra we find X1 =
8.4 and X2 = 4.8.
 Solving for point c we find X1 = 18 and X2 = 0.
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Holiday Meal Turkey Ranch
Substituting these value back into the objective
function we find
Cost = 2X1 + 3X2
Cost at point a = 2(3) + 3(12) = 42
Cost at point b = 2(8.4) + 3(4.8) = 31.2
Cost at point c = 2(18) + 3(0) = 36
The lowest cost solution is to purchase 8.4 pounds
of brand 1 feed and 4.8 pounds of brand 2 feed for a
total cost of 31.2 cents per turkey.
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Holiday Meal Turkey Ranch
Graphical Solution to the Holiday Meal Turkey Ranch
Problem Using the Isocost Approach
X2
–
Feasible Region
Pounds of Brand 2
20 –
15 –
10 –
5–
Figure 7.11
(X1 = 8.4, X2 = 4.8)
0 |–
|
5
|
|
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10
15
20
Copyright ©2012 Pearson Education, Inc. publishing as Prentice HallPounds of Brand 1
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25 X1
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Holiday Meal Turkey Ranch
Solving the Holiday Meal Turkey Ranch Problem
Using QM for Windows
Program 7.3
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Holiday Meal Turkey Ranch
Excel 2010 Spreadsheet for the Holiday Meal Turkey
Ranch problem
Program 7.4A
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Holiday Meal Turkey Ranch
Excel 2010 Solution to the Holiday Meal Turkey
Ranch Problem
Program 7.4B
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Four Special Cases in LP
 Four special cases and difficulties arise at
times when using the graphical approach
to solving LP problems.
 No feasible solution
 Unboundedness
 Redundancy
 Alternate Optimal Solutions
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Four Special Cases in LP
No feasible solution
 This exists when there is no solution to the
problem that satisfies all the constraint
equations.
 No feasible solution region exists.
 This is a common occurrence in the real world.
 Generally one or more constraints are relaxed
until a solution is found.
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Four Special Cases in LP
A problem with no feasible solution
X2
Figure 7.12
8–
–
6–
–
4–
–
2–
–
0–
Region Satisfying
Third Constraint
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2
|
|
4
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6
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8
X1
Region Satisfying First Two Constraints
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7-71
Four Special Cases in LP
Unboundedness
 Sometimes a linear program will not have a
finite solution.
 In a maximization problem, one or more
solution variables, and the profit, can be made
infinitely large without violating any
constraints.
 In a graphical solution, the feasible region will
be open ended.
 This usually means the problem has been
formulated improperly.
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Four Special Cases in LP
A Feasible Region That is Unbounded to the Right
X2
X1 ≥ 5
15 –
X2 ≤ 10
10 –
Feasible Region
5–
X1 + 2X2 ≥ 15
Figure 7.13
0 |–
|
|
|
5
10
15
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|
X1
7-73
Four Special Cases in LP
Redundancy
 A redundant constraint is one that does not
affect the feasible solution region.
 One or more constraints may be binding.
 This is a very common occurrence in the real
world.
 It causes no particular problems, but
eliminating redundant constraints simplifies
the model.
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Four Special Cases in LP
Problem with a Redundant Constraint
X2
30 –
25 –
2X1 + X2 ≤ 30
20 –
Redundant
Constraint
15 –
X1 ≤ 25
10 –
Figure 7.14
5–
0–
X1 + X2 ≤ 20
Feasible
Region
|
|
|
|
|
|
5
10
15
20
25
30
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X1
7-75
Four Special Cases in LP
Alternate Optimal Solutions
 Occasionally two or more optimal solutions
may exist.
 Graphically this occurs when the objective
function’s isoprofit or isocost line runs
perfectly parallel to one of the constraints.
 This actually allows management great
flexibility in deciding which combination to
select as the profit is the same at each
alternate solution.
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Four Special Cases in LP
Example of Alternate Optimal Solutions
X2
8–
7–
6 –A
5–
Optimal Solution Consists of All
Combinations of X1 and X2 Along
the AB Segment
4–
3–
Isoprofit Line for $8
2–
Isoprofit Line for $12
Overlays Line Segment AB
B
Figure 7.15
1 – Feasible
Region
|
|
0–
1
2
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4
5
6
7
8 X1
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Sensitivity Analysis
 Optimal solutions to LP problems thus far have




been found under what are called deterministic
assumptions.
This means that we assume complete certainty in
the data and relationships of a problem.
But in the real world, conditions are dynamic and
changing.
We can analyze how sensitive a deterministic
solution is to changes in the assumptions of the
model.
This is called sensitivity analysis, postoptimality
analysis, parametric programming, or optimality
analysis.
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Sensitivity Analysis
 Sensitivity analysis often involves a series of
what-if? questions concerning constraints,
variable coefficients, and the objective function.
 One way to do this is the trial-and-error method
where values are changed and the entire model is
resolved.
 The preferred way is to use an analytic
postoptimality analysis.
 After a problem has been solved, we determine a
range of changes in problem parameters that will
not affect the optimal solution or change the
variables in the solution.
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7-79
High Note Sound Company
 The High Note Sound Company manufactures
quality CD players and stereo receivers.
 Products require a certain amount of skilled
artisanship which is in limited supply.
 The firm has formulated the following product mix
LP model.
Maximize profit = $50X1 + $120X2
Subject to
2X1
+ 4X2 ≤ 80
3X1
+ 1X2 ≤ 60
(hours of electrician’s
time available)
(hours of audio
technician’s time
available)
X1, X2 ≥ 0
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7-80
High Note Sound Company
The High Note Sound Company Graphical Solution
X2
(receivers)
60 –
–
Optimal Solution at Point a
X1 = 0 CD Players
X2 = 20 Receivers
Profits = $2,400
40 –
a = (0, 20) –
b = (16, 12)
20 –
Isoprofit Line: $2,400 = 50X1 + 120X2
10 –
Figure 7.16
0–
|
|
10
20
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30
40
c = (20, 0)
|
50
|
60
X1
(CD players)
7-81
Changes in the
Objective Function Coefficient
 In real-life problems, contribution rates in the
objective functions fluctuate periodically.
 Graphically, this means that although the feasible
solution region remains exactly the same, the
slope of the isoprofit or isocost line will change.
 We can often make modest increases or
decreases in the objective function coefficient of
any variable without changing the current optimal
corner point.
 We need to know how much an objective function
coefficient can change before the optimal solution
would be at a different corner point.
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Changes in the
Objective Function Coefficient
Changes in the Receiver Contribution Coefficients
X2
40 –
Profit Line for 50X1 + 80X2
(Passes through Point b)
30 –
Old Profit Line for 50X1 + 120X2
(Passes through Point a)
20 –
b
a
Profit Line for 50X1 + 150X2
(Passes through Point a)
10 –
0–
|
10
| c
20
|
|
|
|
30
40
50
60
X1
Figure 7.17
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QM for Windows and Changes in
Objective Function Coefficients
Input and Sensitivity Analysis for High Note Sound
Data Using QM For Windows
Program 7.5A
Program 7.5B
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Excel Solver and Changes in
Objective Function Coefficients
Excel 2010 Spreadsheet for High Note Sound Company
Program 7.6A
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Excel Solver and Changes in Objective
Function Coefficients
Excel 2010 Solution and Solver Results Window for
High Note Sound Company
Figure 7.6B
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Excel Solver and Changes in
Objective Function Coefficients
Excel 2010 Sensitivity Report for High Note Sound
Company
Program 7.6C
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Changes in the
Technological Coefficients
 Changes in the technological coefficients often
reflect changes in the state of technology.
 If the amount of resources needed to produce a
product changes, coefficients in the constraint
equations will change.
 This does not change the objective function, but
it can produce a significant change in the shape
of the feasible region.
 This may cause a change in the optimal solution.
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7-88
Changes in the
Technological Coefficients
Change in the Technological Coefficients for the
High Note Sound Company
(b) Change in Circled
Coefficient
Stereo Receivers
(a) Original Problem
X2
X2
60 –
60 –
40 –
3X1 + 1X2 ≤ 60
20 –a
b
–
0
c
20
|
20 –a
2X1 + 4X2 ≤ 80
|
40
|
X1
–
0
X2
60 –
2 X1 + 1X2 ≤ 60
40 –
Optimal
Solution
(c) Change in Circled
Coefficient
40 –
Still
Optimal
d
2X1 + 4X2 ≤ 80
e |
|
20 30 40
|
|
X1
20 –
16 f
–
0
3X1 + 1X2 ≤ 60
Optimal
Solution
g
c
20
|
2X1 + 5 X2 ≤ 80
|
40
|
X1
CD Players
Figure 7.18
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7-89
Changes in Resources or
Right-Hand-Side Values
 The right-hand-side values of the
constraints often represent resources
available to the firm.
 If additional resources were available, a
higher total profit could be realized.
 Sensitivity analysis about resources will
help answer questions about how much
should be paid for additional resources
and how much more of a resource would
be useful.
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7-90
Changes in Resources or
Right-Hand-Side Values
 If the right-hand side of a constraint is changed,
the feasible region will change (unless the
constraint is redundant).
 Often the optimal solution will change.
 The amount of change in the objective function
value that results from a unit change in one of the
resources available is called the dual price or dual
value .
 The dual price for a constraint is the improvement
in the objective function value that results from a
one-unit increase in the right-hand side of the
constraint.
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7-91
Changes in Resources or
Right-Hand-Side Values
 However, the amount of possible increase in the
right-hand side of a resource is limited.
 If the number of hours increased beyond the
upper bound, then the objective function would no
longer increase by the dual price.
 There would simply be excess (slack) hours of a
resource or the objective function may change by
an amount different from the dual price.
 The dual price is relevant only within limits.
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Changes in the Electricians’ Time Resource
for the High Note Sound Company
X2
(a)
60 –
40 –
25 –
20 –
Constraint Representing 60 Hours of
Audio Technician’s Time Resource
a
–
0
b
Changed Constraint Representing 100 Hours
of Electrician’s Time Resource
|
c
20
|
|
|
40
50
60
X1
Figure 7.19
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7-93
Changes in the Electricians’ Time Resource
for the High Note Sound Company
X2
(b)
60 –
40 –
Constraint Representing 60 Hours of
Audio Technician’s Time Resource
20 –
15 –
Changed Constraint Representing 60 Hours
of Electrician’s Time Resource
a
b
–
0
c
|
|
|
|
20
30
40
60
X1
Figure 7.19
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7-94
Changes in the Electricians’ Time Resource
for the High Note Sound Company
X2
(c)
60 –
Changed Constraint Representing 240 Hours
of Electrician’s Time Resource
40 –
Constraint
Representing
60 Hours of Audio
Technician’s
Time Resource
20 –
–
0
|
|
|
|
|
|
20
40
60
80
100
120
X1
Figure 7.19
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QM for Windows and Changes in
Right-Hand-Side Values
Sensitivity Analysis for High Note Sound Company
Using QM for Windows
Program 7.5B
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7-96
Excel Solver and Changes in
Right-Hand-Side Values
Excel 2010 Sensitivity Analysis for High Note Sound
Company
Program 7.6C
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7-97
Copyright
All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted, in
any form or by any means, electronic, mechanical,
photocopying, recording, or otherwise, without the prior
written permission of the publisher. Printed in the United
States of America.
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